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Network Models Part 1
Network Models
Part 1
Network Models
• There is a many of operation research situation is modeled and solved as network ( nodes can connected by branches)
• There are five network models algorithms1- Minimal spanning tree2- shortest-route algorithms3- maximum-flow algorithms4- minimum cost capacitated network algorithms5- Critical path( CPM) algorithms
Network Models (CONT.)
1- Design of an offshore gas pipeline network connecting wellheads in gulf of Mexico to an inshore delivery points.; the objective of the model is minimize the cost constructing the pipeline.
• The situation represented as Minimal spanning tree.
2- Determination of the shortest route between two cities in a network of roads.
• This situation is shortest-route algorithms
Network Models (CONT.)
3- determination the maximum capacity (in ton per year) of a coal slurry pipeline network
• This situation is maximum flow algorithms
4- determination of the minimum-cost flow schedule from oil field to refineries through a pipeline network.
• This situation is minimum-cost capacitated network algorithms
Network Models (CONT.)
5- determination the time schdule (start and completion date) for activities
• This situation is (CPM) algorithms
Network definitions
• A network consist of set of nodes linked by arcs ( or branches)
• The notion for describing a network is (N, A) where:
– N is set of nodes– A set of arc
Network definitions (cont.)
• Example
• Flow : the amount sent from node i to node j, over an arc that connects them.
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N ={ 1,2,3,4,5}
A={(1,2), (1,3),(2,3),(2,5),(3,4),(3,5),(4,2),(4,5)}
Network definitions (cont.)
• Directed/undirected arcs : • when flow is allowed in one direction the arc is
directed; (that means allow positive flow in one direction and zero flow in the opposite direction)• When flow is allowed in two directions, the arc is
undirected.
• Path : sequence of distinct arcs that join two nodes through other nodes regardless of the direction of flow in each arcs• The nodes are said to be connected if there is a path
between them.8
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• Cycle : a path starting at a certain node and returning to the same node without using any arc twice. (or connects a node to itself through other nodes)
Example:
– (2,3),(3,5),(5.2) form of loop– Cycle is directed if it consists of directed path (2,3),
(3,4) and( 4,2)
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Network definitions (cont.)
• Tree : is connected network that may involve only a subset of all nodes of network without cycle.
• Spanning tree : a tree that connects all the nodes in a network with no cycle( it consists of n -1 arcs).
Network definitions (cont.)
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5Tree Spanning
Tree
Minimal Spanning tree
• It deals with linking the nodes of network, directly or indirectly, using shortest length of connecting branches.
• The typical application occurs in construction of paved roads that link several towns.
Minimal Spanning tree
• The step of procedure are given as follows:– Let N={ 1,2,…n} set of nodes– Ck= set of nodes that have been permanently connected
at iteration K– Ck`= set of nodes as yet to be connected permanently.
• Step 0: set C0= 0, C0`=N• Step 1: start with any node I; set C1={i}, C1`=N-{i}• General step: selected node j in unconnected set
Ck-1` that yield in shortest arcs to a node in the connected set . Link j permanently to Ck-1 and remove it from Ck-1`
- If the set of unconnected nodes is empty stop. Otherwise set k=K+1 and repeat the step
Example (cont.)
• Midwest TV cable company is in the process of providing cable service to five new housing development service areas.
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Example (cont.)
• The algorithms start at node 1
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Example (cont.)
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Example (cont.)
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Example (cont.)
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Example (cont.)
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Example (cont.)
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Example (cont.)
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Example (cont.)
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C3 `
Example (cont.)
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Example (cont.)
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Example (cont.)
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C4 `
Example (cont.)
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Example (cont.)
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Example (cont.)
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Example (cont.)
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Alternate links
Example (cont.)
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Alternate links
Example (cont.)
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Example (cont.)
• Summery of solution
iteration Minimum distance connecting arc
distance Add arc to tree? Cumulative tree distance
1 (1,2) 1 yes 1
2 (2,5) 3 yes 4
3 (2,4) 4 yes 8
4 (4,6) 3 yes 11
5 (4,3) 5 yes 16
Example 2
• Apply minimal spanning tree
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Solution
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Solution (cont.)
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Solution (cont.)
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Example 3
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Solution
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Solution (cont.)
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Solution (cont.)
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Solution (cont.)
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Solution (cont.)
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Solution (cont.)
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Solution (cont.)
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Solution (cont.)
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Solution (cont.)
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Solution (cont.)
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Shortest- Route problem
• The shortest route problem determines the shortest route between a source and destination.
• There are two algorithms to solve shortest-route problems:
• 1- Dijkstra’s algorithm that design to determine the shortest routes between the source node every other node in the network
• 2- Floyd’s algorithms is general because it allow the determination of the shortest route between any two node in network
Dijkstra’a algorithm
Step0: label the source node(node1) with the permanent label [0,--]. Set i=1
Stepi= (a) compute the temporary labels[ui+dij,i] for each node j that can be reached through node i. provided j is not permanently label. If node j is already label with [uj,k] through another node k and if ui+dij< uj, replace [uj,k] with [ ui+ dij, i]
(b) if all node have premanent label stop. Otherwise select the label [Ur,s] having the shortest distance (=ui) among all temporary label. Set i=r and repeat step i
Example
• The figure give the route and their length in miles between city 1 and four other cities. Determine the shortest route between city 1 and each of the remaining four cities.
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Example(cont.)
• Iteration 0: assign permanent label [0,--] to node 1• Iteration 1: node 2 and 3 can be reached from (the last
permanent labeled) node 1 thus the list labeled node (temporary and permanent) becomes
• For both two temporary label[100,1] and [30,1] node 3 is smallest distance so, status of node 3 is changed to permanent
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Node label status
1 [0,--] permanent
2 [0+100, 1] temporary
3 [0+30,1] temporary
Example(cont.)
• Iteration2: node 4, and 5 can be reached from node 3 and the list labeled node becomes:
• node 4 is smallest distance so from the temporaries list. so, status of node 4 is changed to permanent
Node label status
1 [0,--] permanent
2 [100, 1] temporary
3 [30,1] Permanent
4 [30+10,3]=[40,3] temporary
5 [30+60,3]=[90,3] temporary
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Example(cont.)
• Iteration 3: node 2 and 5 can be reached from node4. the list of labeled is updated as
• Node 2 is permanent
Node label status
1 [0,--] permanent
2 [40+12,4]=[55,4] temporary
3 [30,1] Permanent
4 [40,3] Permanent
5 [90,3] or [40+50,4] temporary
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Example(cont.)
• Iteration 4: only node 3 can be reached from node 2, the node 3 is permanent , so the new list remain the same
• Because node 5 is not lead to other node, it is status will convert to permanent
Node label status
1 [0,--] permanent
2 [55,4] permanent
3 [30,1] Permanent
4 [40,3] Permanent
5 [90,3] or [40+50,4] temporary
Example(cont.)
• The process ends
• The shortest route between node1 and node2 is:• (2) [55,4](4) [40,3](3) [30,1](1)• So the disired route is 1342 with total length 55
miles
Node label status
1 [0,--] permanent
2 [55,4] permanent
3 [30,1] Permanent
4 [40,3] Permanent
5 [90,3] or [40+50,4] Permanent
Floyd’s algorithm
• Step0: define starting distance matrix D0 and node sequence matrix S0. the diagonal elements are marked with(-). Set k=1
• General step k: define row k and column as pivot row and pivot column. Apply the triple operation to each element dij in Dk-1. if the condition:
Dik+dkj<dij Is satisfied, make the following changes:• (a) creat Dk by replacing dij in Dk-1 with dik+dkj• (b) create Sk by repacing sij in sk-1 with k. set k=k+1 and
repeat step k.
Floyd’s algorithm (cont.)
• Step k : if the sum elements on the pivot row and povot coumn is smaller thanassociated intersection elements, the it is optimal to replace the intersection distance by the sum of pivot distance.
• After n step, it can determine the shortest route by using the following rules:
• 1- from D dij gives the shortest distance
• 2- from S determine the intermediate node.
Example
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D0 s0
Example (cont.)
• K=1
• We highlight the first column and first row of the Distance matrix and compare all other items with the sum of the items highlighted in the same row and column.
• If the sum is less than the item then it should be replaced with the sum.
Example (cont.)
• When (-) is involved we leave the item.
D0 S0
Example (cont.)
• 10+3=13 is less than ∞
So change
D0 S0
Example (cont.)
• 10+3=13 is less than ∞
So change
D0 S0
Example (cont.)
• When ∞is involved we leave the item.
D0 S0
Example (cont.)
• When ∞is involved we leave the item.
D0 S0
Example (cont.)
• 3+10=13 less than ∞,
So change
D0 S0
Example (cont.)
• When (-) is involved we leave the item.
D0 S0
Example (cont.)
• When ∞is involved we leave the item.
D0 S0
Example (cont.)
When ∞is involved we leave the item.
D0 S0
Example (cont.)
• When ∞is involved we leave the item.
D0 S0
Example (cont.)
• When ∞is involved we leave the item.
D0 S0
Example (cont.)
• When (-) is involved we leave the item.
D0 S0
Example (cont.)
• When ∞is involved we leave the item.
D0 S0
Example (cont.)
• When ∞is involved we leave the item.
D0 S0
Example (cont.)
• When ∞is involved we leave the item.
D0 S0
Example (cont.)
• When ∞is involved we leave the item.
D0 S0
Example (cont.)
• When ∞is involved we leave the item.
D0 S0
Example (cont.)
Example (cont.)
• We have now completed one iteration. We rename the new matrices:
D1 S1
Example (cont.)
• Set k=2
• We highlight the second column and second row of the Distance matrix and compare all other items with the sum of the items highlighted in the same row and column.
• If the sum is less than the item then it should be replaced with the sum.
Example (cont.)
• When (-) is involved we leave the item.
D1 S1
Example (cont.)
• 3+13=16 Not Less than 10
D1 S1
Example (cont.)
• 3+5=8 less than ∞
So change
D1 S1
Example (cont.)
• 3+5=8 less than ∞
So change
D1 S1
Example (cont.)
• When ∞is involved we leave the item.
D1 S1
Example (cont.)
• 3+13=16 Not less than 10
D1 S1
Example (cont.)
• When (-) is involved we leave the item.
D1 S1
Example (cont.)
• 5+13=18 Not less than 6
D1 S1
Example (cont.)
• When ∞is involved we leave the item.
D1 S1
Example (cont.)
• 3+5=8 less than ∞
So change
D1 S1
Example (cont.)
• 3+5=8 less than ∞
So change
D1 S1
Example (cont.)
• 13+5=18 Not less than 6
D1 S1
Example (cont.)
• When (-) is involved we leave the item.
D1 S1
Example (cont.)
• When ∞is involved we leave the item.
D1 S1
Example (cont.)
• When ∞is involved we leave the item.
D1 S1
Example (cont.)
• When ∞is involved we leave the item.
D1 S1
Example (cont.)
• When ∞is involved we leave the item.
D1 S1
Example (cont.)
• When ∞is involved we leave the item.
D1 S1
Example (cont.)D1 S1
Example (cont.)
• We have now completed two iteration. We rename the new matrices:
D2S2
Example (cont.)
• Set k=3
• We highlight the third column and third row of the Distance matrix and compare all other items with the sum of the items highlighted in the same row and column.
• If the sum is less than the item then it should be replaced with the sum.
Example (cont.)D2 S2
Example (cont.)D2 S2
Example (cont.)D2 S2
Example (cont.)
• We have now completed third iteration. We rename the new matrices:
D3 S3
Example (cont.)
• Set k=4
• We highlight the fourth column and fourthrow of the Distance matrix and compare all other items with the sum of the items highlighted in the same row and column.
• If the sum is less than the item then it should be replaced with the sum.
Example (cont.)
D3 S3
Example (cont.)
D3 S3
Example (cont.)
• We have now completed fourth iteration. We rename the new matrices:
D4 S4
Example (cont.)
• Set k=5
• We highlight the fifth column and fifth row of the Distance matrix and compare all other items with the sum of the items highlighted in the same row and column.
• If the sum is less than the item then it should be replaced with the sum.
Example (cont.)
• No further improvement are possible in this iteration, D5,S5 are the same D4 and S4
D4 S4
Example (cont.)
• Shortest distance is d15 =12
• Associated route: recall segment(I,j) if Sij=J is direct link otherwise they link through intermediate node.
D4 S4
Example (cont.)
• S15= 4 ≠ 5 so. The initial link is 145
• Now, s14=2. is not direct link and 14 must replaced with 124, so the road from 1 to 5 will be change to 1245.
• Now s12=2, s24=4,s45=5. the route 1245 need no further dissecting and the process end
